If $f_0\ne0$ or if $f_0=0$ and the $f_1,\dotsc,f_n$ are not linearly independent, then the answer is trivial:

In this case there is another functional, say $f_1$, which is in the span of the remaining ones, say $f_1=\sum_{k\ne1}\lambda_kf_k$: Extend the $f_k$ to $F_k$ $(k=2,\dotsc,n)$ and put $F_1=\sum_{k\ne1}\lambda_kF_k$.

In the remaining case, $f_0=0$ and $f_1,\dotsc,f_n$ being linearly independent, the answer to your question is obviously positive if and only if $F_0=0$ (because $\sum_{k=1}^n\lambda_kF_k=F_0$ implies by restriction to the subspace $\lambda_1=\dotsc=\lambda_n=0$).

If $f_0\ne0$ or if $f_0=0$ and the $f_1,\dotsc,f_n$ are not linearly independent, then the answer is trivial:

In this case there is another functional, say $f_1$, which is in the span of the remaining ones, say $f_1=\sum_{k\ne1}\lambda_kf_k$ with $\lambda_0\ne0$: Extend the $f_k$ to $F_k$ $(k=2,\dotsc,n)$ and put $F_1=\sum_{k\ne1}\lambda_kF_k$.

In the remaining case, $f_0=0$ and $f_1,\dotsc,f_n$ being linearly independent, the answer to your question is obviously positive if and only if $F_0=0$ (because $\sum_{k=1}^n\lambda_kF_k=F_0$ implies by restriction to the subspace $\lambda_1=\dotsc=\lambda_n=0$).